Integration by Substitution

Transcription

1 Integration by Substitution MATH 151 Calculus for Management J. Robert Buchanan Department of Mathematics Fall 2018

2 Objectives After this lesson we will be able to use the method of integration by substitution to find the integrals of more complicated functions.

3 Power Rule Recall the Power Rule for integration, x r dx = 1 r + 1 x r+1 + C if r 1.

4 Power Rule Recall the Power Rule for integration, x r dx = 1 r + 1 x r+1 + C if r 1. From the simple power rule we can develop a more generally applicable integration formula.

5 Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4.

6 Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4. Notice that the derivative of x is 3x 2. f (x) may be the result of applying the chain rule for derivatives to some function of x

7 Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4. Notice that the derivative of x is 3x 2. f (x) may be the result of applying the chain rule for derivatives to some function of x d [ (x 3 + 2) 5] = 5(x 3 + 2) 4 (3x 2 ) dx [ ] d 1 dx 5 (x 3 + 2) 5 = 3x 2 (x 3 + 2) 4

8 Example Suppose we must find an antiderivative of f (x) = 3x 2 (x 3 + 2) 4. Notice that the derivative of x is 3x 2. f (x) may be the result of applying the chain rule for derivatives to some function of x d [ (x 3 + 2) 5] = 5(x 3 + 2) 4 (3x 2 ) dx [ ] d 1 dx 5 (x 3 + 2) 5 = 3x 2 (x 3 + 2) 4 Therefore 3x 2 (x 3 + 2) 4 dx = 1 5 (x 3 + 2) 5 + C.

9 Substitution Consider the indefinite integral: (x 3 + 2) 4 3x 2 dx

10 Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du

11 Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du = u 4 du

12 Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du = u 4 du = 1 5 u5 + C

13 Substitution Consider the indefinite integral: (x } 3 {{ + 2 } ) 4 3x }{{ 2 dx } u du = u 4 du = 1 5 u5 + C = 1 5 (x 3 + 2) 5 + C

14 General Power Rule Theorem If u is a differentiable function of x, then u n du dx dx = u n du = 1 n + 1 un+1 + C if n 1.

15 General Power Rule Theorem If u is a differentiable function of x, then u n du dx dx = u n du = 1 n + 1 un+1 + C if n 1. Remark: in some integrals we may be able to substitute the symbol u for a function and then carry out an elementary integration. This process is known as integration by substitution.

16 Integration with Function Notation k g (x) dx = k g(x) + C (g(x)) r g 1 (x) dx = r + 1 (g(x))r+1 + C (if r 1) g (x) dx = ln g(x) + C g(x) e g(x) g (x) dx = e g(x) + C.

17 Integration with u-substitution Suppose u is a differentiable function of x, then k du = ku + C u r 1 du = r + 1 ur+1 + C (if r 1) 1 du = ln u + C u e u du = e u + C.

18 Guidelines for Integration by Substitution 1. Let u be a function of x (usually part of a composition of functions in the integrand). 2. Solve for x and dx in terms of u and du. 3. Convert the integral in variable x into an integral in variable u. 4. After integrating, rewrite the antiderivative as a function of x. 5. Check the answer by differentiating. f (g(x)) g (x) dx = f (u) du = f (u) + C = f (g(x)) + C

19 Example (1 of 4) Evaluate the indefinite integral (3x 2 + 6)(x 3 + 6x) 2 dx.

20 Example (1 of 4) Evaluate the indefinite integral (3x 2 + 6)(x 3 + 6x) 2 dx. If we let u = x 3 + 6x then du = (3x 2 + 6) dx. (3x 2 + 6)(x 3 + 6x) 2 dx = u 2 du = 1 3 u3 + C = 1 ( 3 x 3 + 6x) + C 3

21 Example (2 of 4) Evaluate the indefinite integral 2(3x 4 + 1) 2 dx.

22 Example (2 of 4) Evaluate the indefinite integral 2(3x 4 + 1) 2 dx. If we let u = 3x then du = 12x 3 dx which cannot be made to equal 2 dx, thus here we must expand the product and then integrate without substitution. ( 2(3x 4 + 1) 2 dx = 18x x 4 + 2) dx = 2x x 5 + 2x + C

23 Example (3 of 4) Evaluate the indefinite integral x 2 e 2x 3 dx.

24 Example (3 of 4) Evaluate the indefinite integral x 2 e 2x 3 dx. If we let u = 2x 3 then du = 6x 2 dx 1 6 du = x 2 dx. Making the substitution we get x 2 e 2x 3 dx = 1 6 eu du = 1 6 eu + C = 1 6 e 2x 3 + C.

25 Example (4 of 4) Evaluate the indefinite integral (ln x) 3 dx. x

26 Example (4 of 4) Evaluate the indefinite integral (ln x) 3 dx. x If we let u = ln x then du = 1 x dx. Making the substitution we get (ln x) 3 dx = (ln x) 3 1 x x dx = u 3 du = 1 4 u4 + C = 1 4 (ln x)4 + C.

27 Propensity to Consume As the income level of a family increases, the family tends to save more and spend less than 100% of their income on necessities. The rate of change of consumption with respect to income is called the marginal propensity to consume. Example Suppose that a family of four with a total income of $25, 000 will spend 100% of their income on necessities. As the income of the family increases above $25, 000 the marginal propensity to consume is dq dx = 0.94 (x 24, 999) 0.06.

28 Example 1. Find the consumption Q as a function of income x. 2. Find the income saved and the income consumed if x = $75, 000.

29 Example 1. Find the consumption Q as a function of income x. Q(x) = 0.94 dx (x 24, 999) 0.06 = 0.94 u 0.06 du = u C Q(x) = (x 24, 999) C Q(25, 000) = 25, 000 = (25, , 999) C Q(x) = (x 24, 999) , Find the income saved and the income consumed if x = $75, 000.

30 Example 1. Find the consumption Q as a function of income x. Q(x) = 0.94 dx (x 24, 999) 0.06 = 0.94 u 0.06 du = u C Q(x) = (x 24, 999) C Q(25, 000) = 25, 000 = (25, , 999) C Q(x) = (x 24, 999) , Find the income saved and the income consumed if x = $75, 000. Q(75, 000) = (75, , 999) , 999 = 51, 123 consumed and 75, , 123 = 23, 877 saved.

31 Demand Function Suppose the quantity demanded x is a function of the price per unit p for a certain product. If dx dp = 300 (0.03p 1) 3 and x = 10, 000 when p = $105 find the demand function x = f (p).

32 Solution dx dp = 300 (0.03p 1) 3 If u = 0.03p 1 then du = dp and 300 x = (0.03p 1) 3 dp 100 = u 3 du = u 3 du f (p) = 5000u C = (0.03p 1) 2 + C f (105) = 5000 (0.03(105) 1) 2 + C = f (p) = 5000 (0.03p 1)